Affine plane curves with constant curvature

Question

Question

I want to solve this differential equation for $P : \mathbb{R} \to \mathbb{A}^2$, a plane affine curve.

$ P'''(t) = \frac{P'(t)}{t^2}$

Someone recognize this equation? Is a famous curve? Is algebraic? I think it will be a curve invariant under the action of a 1-parameter subgroup of the affine group.

Thanks in advance.

Summary

I considered again the results found in

Affine arc length

I started studying some specials curves:

1. $\omega^1 = 0$. These are points and are invariant under the action of a 4-parameter subgroup of the affine group. For example for $x^2+y^2=0$ $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & c & d \\ \end{bmatrix} $$

2. $\omega_1^2 = 0$. These are straight lines and are invariant under the action of a 4-parameter subgroup of the affine group. For example for $y = 0$ $$ \begin{bmatrix} 1 & 0 & 0 \\ x & a & b \\ 0 & 0 & d \\ \end{bmatrix} $$

3. $\omega_2^1 = 0$. These are parabolas and are invariant under the action of a 2-parameter subgroup of the affine group. For example for $y = x^2$ $$ \begin{bmatrix} 1 & 0 & 0 \\ x & a & 0 \\ x^2 & 2ax & a^2 \\ \end{bmatrix} $$

4. $k = 0$. These are ellipses and hyperbolas and are invariant under the action of a 1-parameter subgroup of the affine group. For example for $x^2+y^2=1$ $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta \\ 0 & -\sin\theta & \cos\theta \\ \end{bmatrix} $$ and for $xy=1$ $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & 1/a \\ \end{bmatrix} $$

5. $k = K$. I got the differential equation $ P'''(t) = \frac{P'(t)}{t^2}$, but I don't know if it is a well-known curve and especially if it is still algebraic.

Answer 1

I relaxed a bit the conditions about the parametrization.

I searched for P such that $P''' = \alpha P'' + \beta P'$ with $\alpha, \beta$ constant. So $\lambda^{-1} = \sqrt{\alpha + \frac{2}{9}\beta^2}$. In particular I can choose $\lambda = 1, i$. For example, if I choose $\lambda =1$, I get $\alpha = 1 - \frac{2}{9}\beta^2$ and $\kappa = \frac{\beta}{3}$ and the equation $P''' - \beta P'' - (1 - \frac{2}{9}\beta^2)P' = 0$ and that's easy to solve. So I have found:

1. $d\sigma^2 = 1$

   • $P = (e^{\frac{3k-\sqrt{k^2+4}}{2}t}, e^{\frac{3k+\sqrt{k^2+4}}{2}t})$ for $|k| \neq \frac{1}{\sqrt{2}}$
   • $P = (t, e^{\pm\frac{3}{\sqrt{2}}t})$ for $|k| = \frac{1}{\sqrt{2}}$

2. $d\sigma^2 = -1$

   • $P = (e^{\frac{3}{2}kt}\cos{\sqrt{4-k^2}t}, e^{\frac{3}{2}kt}\sin{\sqrt{4-k^2}t}$) for $|k| < 2$
   • $P = (e^{\pm3t}, te^{\pm3t})$ for $|k| = 2$
   • $P = (e^{\frac{3k-\sqrt{k^2-4}}{2}t}, e^{\frac{3k+\sqrt{k^2-4}}{2}t})$ for $|k| > 2$

So every curve with constant affine curvature is congruent to one of

1. $P = (t, t^{\mu})$ with $\mu \ne \frac{1}{2}, 2$ (from 1.1 and 2.3)
2. $P = (t, e^{\pm t})$ (from 1.2)
3. $P = (e^t, \pm te^t)$ (from 2.2)
4. $P = (e^{\lambda t}\cos t, e^{\lambda t}\sin t) $ (from 2.1)

The algebraic one are only of type 1 with $\mu \in \mathbb{Q}$ (or type 4 with $\lambda = 0$)